Construct an isosceles triangle whose base is 6 cm and altitude 4 cm. Then construct another triangle whose sides are `3/4`times the corresponding sides of the isosceles triangle.

#### Solution

Let ABC be an isosceles triangle with AB = AC, base BC = 6 cm and altitude AD = 4 cm.

A ΔA'BC', whose sides are 34 times the sides of ΔABC, can be drawn using the following steps:

1. Draw a line segment BC of 6 cm. Draw arcs of the same radius on both sides of the line segment while taking points B and C as its centre. Let these arcs intersect each other at O and O'. Join OO'. Let OO' intersect BC at D

2. Taking D as centre, draw an arc of 4 cm radius which cuts the line segment OO' at point A. An isosceles ΔABC is formed, having altitude (AD) as 4 cm and base (BC) as 6 cm.

3. Draw a ray BX making an acute angle with line segment BC on the opposite side of vertex A.

4. Locate 4 points (as 4 is greater between 3 and 4) B_{1}, B_{2}, B_{3} and B_{4} on BX such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}.

5. Join CB_{4} and draw a line through B_{3} parallel to CB_{4} to intersect line segment BC at point C'.

6. Draw a line through C' parallel to AC intersecting line segment AB at A'.

Thus, ΔA'BC' is the required triangle